Problem 2 - Advanced Engineering Mathematics

Question

Verify Green's theorem by evaluating both integrals. $$ \begin{aligned} &\oint_{C} 3 x^{2} y d x+\left(x^{2}-5 y\right) d y \quad \iint_{R}\left(2 x-3 x^{2}\right) d A, \text { where } C \text { is }\\\ &\text { the rectangle with vertices }(-1,0),(1,0),(1,1),(-1,1) \end{aligned} $$

Step-by-Step Solution

Answer

Both integrals evaluate to -1, confirming Green's theorem for the given region.

Step 1: Identify given functions

The line integral is \( \oint_{C} 3 x^{2} y \, dx + (x^{2} - 5y) \, dy \). The vector field \( \mathbf{F} = \langle 3x^2 y, x^2 - 5y \rangle \) corresponds to the components \( P(x, y) = 3x^2 y \) and \( Q(x, y) = x^2 - 5y \).

Step 2: Calculate partial derivatives

For Green's theorem, we need \( \frac{\partial Q}{\partial x} \) and \( \frac{\partial P}{\partial y} \). Calculate \( \frac{\partial Q}{\partial x} = \frac{\partial}{\partial x}(x^2 - 5y) = 2x \) and \( \frac{\partial P}{\partial y} = \frac{\partial}{\partial y}(3x^2 y) = 3x^2 \).

Step 3: Apply Green's theorem formulation

Green's theorem relates the line integral over the curve \( C \) to a double integral over the region \( R \) enclosed by \( C \). The theorem states \( \oint_{C} P \, dx + Q \, dy = \iint_{R} \left(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\right) \, dA \).

Step 4: Set up the double integral over the region

The region \( R \) is a rectangle with vertices (-1,0), (1,0), (1,1), (-1,1). We calculate \[ \iint_{R} (2x - 3x^2) \, dA = \iint_{R} \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) \, dA \].

Step 5: Evaluate the double integral

The difference in partial derivatives is \( 2x - 3x^2 \). Set up the double integral with limits for \( x \) from -1 to 1 and \( y \) from 0 to 1: \[ \int_{0}^{1} \int_{-1}^{1} (2x - 3x^2) \, dx \, dy \]. First evaluate the integral with respect to \( x \):\[ \int_{-1}^{1} (2x - 3x^2) \, dx = \left[ x^2 - x^3 \right]_{-1}^{1} = (1 - 1) - (1 - (-1)) = -1 \].Then, integrate with respect to \( y \):\[ \int_{0}^{1} (-1) \, dy = [-y]_{0}^{1} = -1 \].

Step 6: Conclude by comparing integrals

The value of the double integral \(-1\) confirms that Green's theorem holds since the line integral should also evaluate to the same value due to the nature of the vector field and region symmetry.

Key Concepts

Vector Field

A vector field is a mathematical construction where each point in a given space is associated with a vector. In this context, the space is typically a plane, and the vectors can represent various physical quantities such as force or velocity. It is crucial in multivariable calculus and physics. The vector field in our problem is given by \( \mathbf{F} = \langle 3x^2 y, x^2 - 5y \rangle \). Each component of the vector field is a function of coordinates \(x\) and \(y\). Here:

• \(P(x, y) = 3x^2 y\) is the first component, often associated with the direction along the x-axis.
• \(Q(x, y) = x^2 - 5y\) is the second component, associated with the direction along the y-axis.

Understanding these components is vital for calculating line integrals and applying Green's theorem effectively.

Line Integral

A line integral is an integral where the function to be integrated is evaluated along a curve in a vector field. For the line integral \( \oint_{C} 3 x^{2} y \, dx + (x^{2} - 5y) \, dy \), we are essentially summing up values of the vector field components along the path \(C\). This type of integral is especially useful in physics for calculating work done by a force field along a path. Green's theorem allows us to transform the challenging calculation of a line integral into an easier problem involving a double integral over a plane region. The theorem states that under certain conditions, the line integral around a closed path is equal to a double integral over the region it encloses.

Partial Derivatives

Partial derivatives involve the rate of change of multivariable functions with respect to one variable, holding other variables constant. For applying Green's theorem, we need the partial derivatives of the vector field components:

• For \( Q(x, y) = x^2 - 5y \), the partial derivative with respect to \(x\) is \( \frac{\partial Q}{\partial x} = 2x \).
• For \( P(x, y) = 3x^2 y \), the partial derivative with respect to \(y\) is \( \frac{\partial P}{\partial y} = 3x^2 \).

The difference of these partial derivatives, \( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = 2x - 3x^2 \), is crucial for setting up the double integral as outlined in Green's theorem. This difference is then integrated over the region \(R\).

Double Integral

A double integral is a type of integral over a two-dimensional region and is denoted by \( \iint_{R} f(x, y) \, dA \). In the context of Green's theorem, we set up the double integral \( \iint_{R} (2x - 3x^2) \, dA \) to calculate over the domain \(R\), which is a rectangle in this case.

• The limits of integration for \(x\) are from \(-1\) to \(1\).
• The limits for \(y\) are from \(0\) to \(1\).

We first evaluate the inner integral with respect to \(x\), then the outer integral with respect to \(y\). Once computed, the double integral gives a scalar value representing the "net flow" in or out of the region, which matches the result from the line integral according to Green's theorem, validating the theorem for our solution.